Show that the cubic equation has one positive and two negative roots. How?

Use Descartes' rule of signs.

You're expected to bracket the roots. If you find a, b with f(a) < 0, f(b) > 0, then you know f has a root between a and b.

So if, for example, you could show f(-10)<0, f(-9)>0 amd f(-8) < 0, you'd know there was a root between -10 and -9, and a root between -9 and -8.
Similarly, if you showed f(10) < 0 and f(11) > 0, you'd know there was a root between 10 and 11.

(Those numbers won't work - you'll have to find some that do by trial and error).

Instead of trial and error, it might help to sketch the graph, calculating a few key points:

Maximum point,
Minimum point,
Where the line meets the y-axis.

With these three data, you can see how many roots are negative and how many are positive.

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Show that the equation x^3-12x-7.2=0 has one positive and two negative roots. Obtain the positive root to 3 significant figures using the Newton-Raphson process.
How do i do the blue part. I am clueless.

Know that the cubic equation $ax^3 + bx^2 + cx + d = 0$ has roots $\alpha, \beta, \gamma$, and that:

$\\\alpha + \beta + \gamma = -\frac{b}{a}[br]\\ \beta\gamma + \gamma\alpha + \alpha\beta = \frac{c}{a}[br]\\ \alpha\beta\gamma = -\frac{d}{a}$

By substituting in the coefficients, you can prove the negativity or otherwise of the roots.

That's been removed from most syllabuses, sadly.

I guess polynomials altogether will be removed soon.

I guess polynomials altogether will be removed soon.

So I am not allowed to use your method?

Of course you can, but you should know what you're doing.

Otherwise, I would recommend hykim1's method. Foolproof, really.

Instead of trial and error, it might help to sketch the graph, calculating a few key points:

Maximum point,
Minimum point,
Where the line meets the y-axis.

With these three data, you can see how many roots are negative and how many are positive.